Tutorial on the Quaternions of Hamilton

- Michele_Laino

Tutorial on the Quaternions of Hamilton

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- schrodinger

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- Michele_Laino

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- Astrophysics

Thank you Michele, I will give it a read! :)

- Michele_Laino

thanks!! :) @Astrophysics

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## More answers

- oleg3321

is this some sort of new language??? lol

- Michele_Laino

no, the quaternions were introduced by W. R. Hamilton in the far 1833 :) lol

- oleg3321

lol looks very complicated

- Michele_Laino

no, they are not complicated, please study my definitions first @oleg3321

- Astrophysics

Very easy to understand, thanks again!

- Michele_Laino

:) @Astrophysics

- welshfella

Interesting article. Well explained.
Hamilton was Ireland's most famous mathematician . Unfortunately his life was cut short by alcoholism.

- Michele_Laino

Thanks! @welshfella

- DanJS

I remember my calc 3 teacher mentioning those to look at if you got bored..
thanks @Michele_Laino

- Michele_Laino

thanks! :) @DanJS

- arindameducationusc

Thank you @Michele_Laino .
I will definitely give it a read

- Michele_Laino

Thanks! :) @arindameducationusc

- anonymous

quaternions aren't a field, they are a division ring

- Michele_Laino

Quaternions are a field @oldrin.bataku

- anonymous

also you should introduce those other basis matrices \(\mathbf i,\mathbf k\) more naturally -- something like this: consider for \(z=a+bi\) we have: $$\begin{align*}a+bi\mapsto \begin{pmatrix}a+bi&0\\0&a-bi\end{pmatrix}&=\begin{pmatrix}a&0\\0&a\end{pmatrix}+\begin{pmatrix}bi&0\\0&-bi\end{pmatrix}\\&=a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}i&0\\0&-i\end{pmatrix}\\&=a\cdot\mathbf1+b\cdot\mathbf i\end{align*}$$
and then we have that \(\mathbf i\cdot\mathbf j=\mathbf k\)

- anonymous

https://en.wikipedia.org/wiki/Quaternion#Noncommutativity_of_multiplication
>The fact that quaternion multiplication is not commutative makes the quaternions an often-cited example of a strictly skew field.

- Michele_Laino

I wrote that formula:
\[{\mathbf{ij}} = - {\mathbf{ji}} = {\mathbf{k}}\]

- anonymous

https://en.wikipedia.org/wiki/Division_ring
>In abstract algebra, a division ring, also called a skew field, is a ring in which division is possible.
>Division rings differ from fields only in that their multiplication is not required to be commutative. ... Historically, division rings were sometimes referred to as fields, while fields were called “commutative fields”.

- anonymous

yes, I know you did, but you introduced \(\mathbf k\) as a matrix first without any motivation. the reason for its existence is that we have four linearly independent matrices \(\mathbf {1,i,j,ij}\) since the multiplication is bilinear, and then we just call \(\mathbf {ij}=\mathbf k\)

- Michele_Laino

I never worked with non-commutative algebra

- Michele_Laino

I meant I never cited non-commutative algebra

- anonymous

? fields have commutative multiplication by definition, whereas the quaternions \(\mathbb H\) are a non-commutative algebra over \(\mathbb R\) and a division ring, not a field

- Michele_Laino

when I wrote a matrix in M2(C) it is supposed that the multiplication is non-commutative since as you know, the multiplication betwen matrices is non-commutative

- anonymous

lol, yes, clearly, and these matrices do not commute, so they cannot form a field. the quaternions are not a field.

- zzr0ck3r

:) right

- zzr0ck3r

For real man, tuts that are wrong are even worse. This is all simple definition stuff you are getting wrong...Please remove.

- Michele_Laino

More precisely, I have used the term "field", since it is commonly used in mathematics literature, a more appropriate term can be "sfield" which indicate a non-commutative field.
I have supposed that, being the non-commutative characteristic evident, there was no need to use the term "sfield"

- anonymous

field is not commonly used in mathematics literature for a skew field, this isn't the 1800s anymore

- anonymous

it's called a division ring, that is the appropriate, correct, and commonly used term for algebraic structures like this

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